When \(x_1,\ldots,x_i\) are chosen, the average is \(\frac {x_1+\ldots+x_i}{i}\). This is integer if and only if the numerator, i.e. the sum of the chosen terms, is a multiple of \(i\).

Here, consider the following \(\mathrm {dp}\) for each \(i=1,\ldots,N\). (DP stands for Dynamic Programming.)

• \(\mathrm{dp}_{j,k,l}\): the number of ways, modulo \(998244353\), to choose \(k\) terms from the first \(j\) terms of \(A\) such that the remainder by \(i\) of the sum of the chosen terms equals \(l\)

Then the answer is the sum of \(\mathrm{dp}_{N,i,0}\).
For the transitions of \(\mathrm{dp}\), please refer to the sample code. (The \(23\)-rd corresponds to choosing \(a_j\), and the \(24\)-th corresponds not choosing.)

Sample code (C++)

#include <bits/stdc++.h>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;
using mint = modint998244353;

int main() {
    
	int N;
	cin>>N;
	
	vector<int> a(N);
	for(int i=0;i<N;i++)cin>>a[i];
	
	mint ans = 0;
	
	for(int i=1;i<=N;i++){
		vector dp(N+1,vector(i+1,vector<mint>(i,0)));
		dp[0][0][0] = 1;
		for(int j=0;j<N;j++){
			for(int k=0;k<=i;k++){
				for(int l=0;l<i;l++){
					dp[j+1][k][l] += dp[j][k][l];
					if(k!=i)dp[j+1][k+1][(l+a[j])%i] += dp[j][k][l];
				}
			}
		}
		ans += dp[N][i][0];
	}
	
	cout<<ans.val()<<endl;
	
	return 0;
}